Difference between two batch scripts

Question

I got a batch script with a block of code as:

@echo off
setlocal EnableDelayedExpansion
rem US locale, ie: 'Thu 12/02/2015'
for /F "tokens=2 delims=/ " %%m in ("%date%") do set /A "n=(3*((1%%m)%%100-1))"
echo %n%
pause

and while I was trying to understand it, I landed writing it myself as:

@echo off
setlocal EnableDelayedExpansion
rem US locale, ie: 'Thu 12/02/2015'
for /F "tokens=2 delims=/ " %%m in ("%date%") do set /A "n=3*(%%m-1)"
echo %n%
pause

Since both returns 33 as output, can anyone please help me understand the logic behind "n=(3*((1%%m)%%100-1))" and the difference between both the blocks.

Stephan · Accepted Answer · 2015-12-02 11:24:17Z

3

3*((1%%m)%%100-1)

%%m is 12 in this example, resulting in

3*((112)%%100-1)

let's get rid of the redundant parantheses and add some spaces for better readabiltiy:

3 * ( 112 %% 100 ) - 1

evaluated:

3 * (     12     ) - 1

%% is the "Modulo" Operator - it gives back the rest when dividing the first number throug the second one (112 Modulo 100 is (1*100) rest 12) (NOTE: if you try this on commandline instead of inside a batchfile, use a single % only)

This seems to be ridiculous, but think, your number isn't 12, but 09. Numbers starting with Zero are handled as octal, but 09(octal) isn't a valid number, so you'll get a Syntax error.

Same walkthrough with 09 instead of 12:

3*((109)%%100-1)
3 * ( 109 %% 100 ) - 1
3 * (      9     ) - 1

The last line without the Modulo trick would be:

3 * (     09     ) - 1

As 09 is not a valid number (there is no 9 in octal numbers), this wouldn't work.

answered Dec 2, 2015 at 11:24

Stephan

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1 Comment

Kuntal Shaw Over a year ago

Thank you so much! That's a great explanation... :)

Magoo · Accepted Answer · 2015-12-02 11:23:37Z

0

It's possible that the digit-group being processed is "08" or "09".

Batch assumes that a leading "0" means that the following string is octal not decimal and objects to 8 or 9 since they are not octal digits.

the method prefixes a "1" amking "108" or "109" which are bot valid decimal numbers that do not start "0" and hence will be interpreted as decimal.

Since 100 is added to each of the numbers 01..12 then you add 100, you need to remove that 100, so %%100 does that (calculate mod-100 = 1..12)

Another way would be to subtract 100, obviously.

(assuming the group is a month. If it's a *day** as it would appear, the same argument applies, but the limits are 01..31 or 1..31.)

answered Dec 2, 2015 at 11:23

Magoo

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